Section 4.5 – Integration by
Substitution
The Chain Rule and Integration


Find F' x  if F  x   x  3x  5 .
f u  u

9
2
u  x  3x  5
2
9
f ' u  9u 8
u'  2 x  3
F '  x   f ' u   u '
 9u8  2 x  3
 9  x  3 x  5   2 x  3
2
8

THUS:
 
2

9 x  3x  5
u
  2 x  3 dx   x
8
du
2
 3x  5   C
9
Integration by Substitution
Let f, g, and u be differentiable functions of x
such that
f  x   g  u  du
dx
Then

f  x  dx   g  u  du
dx dx   g  u  du  G  u   C
where G is an antiderivative of g.
Substitution Guidelines
1. Choose a substitution u = g(x). Usually, it is
best to choose the inner part of a composite
function, such as a quantity raised to a
power.
2. Compute du = g '(x)dx.
3. Rewrite the integral in terms of u. Make sure
every x and dx is no longer in the integral.
4. Find the resulting integral in terms of u.
5. Replace u by g(x) to obtain an antiderivative
in terms of x.
6. If you have a definite integral, make sure
you change the limits of integration to be in
terms of u before you integrate.
Example 1

Find  9 x  3x  5
2
Define u and du:
  2 x  3. dx
8
u  x 2  3x  5
du  2x  dx  3 dx
du   2 x  3 dx
Substitute to replace EVERY x and dx:
 9  x  3x  5  2 x  3 dx   9 u  du
2
8
8
 9  u  du
8
Integrate.
 9  811 u 81  C
Substitute
back to Leave
your answer in
terms of x.
 u9  C
  x  3x  5   C
2
9
Example 2
Find
 xx
2
 7  dx .
Define u and du:
5
u  x2  7
du  2x  dx
1
2 x du  dx
Substitute to replace EVERY x and dx:
 xx
2
Solve
for dx
 7  dx   x  u  dx
5
5
  x u 
5 1
2x

Substitute
back to Leave
your answer in
terms of x.
1
2
 u 
5
du
du
Integrate.
 12  511 u 51  C
 121 u 6  C

1
12
x
2
 7  C
6
Example 3
Find
1
 cos 2 2 x dx
Define u and du:
  sec 2 2x dx
Rewrite
u  2x
du  2  dx
1
2 du  dx
Substitute to replace EVERY x and dx:
 sec
2
 sec u dx
  sec u du
  sec u du
2x dx 
2
2
Substitute
back to Leave
your answer in
terms of x.
1
2
1
2
2
 12  tan u  C
 12  tan 2x  C
Solve for dx
Integrate.`
Example 4

Find x 5 x  1 dx .
u  5x  1
1
5  u  1  x
du  5  dx
1
Solve for
5 du  dx
Define u and du:
dx
Substitute to replace EVERY x and dx:
x
5 x  1 dx 
Integrate.
x
u dx
  x u 15 du
  15  u  1 u 15 du

1
25

1
25

1
25
There is still an x.
Solve the initial
equation of u for x.
12
u

1
u
du



Substitute back to Leave
your answer in terms of x.
32
12
u

u
 du


2
5
u
52
 u
2
3
32
  C   5x  1
2
125
52

2
75
 5 x  1
32
C
Example 5
Find F(x) if F(0.5) = 4 and F '  x   4 x  1 .
Define u
and du:
u  4x 1
du  4  dx
1
4 du  dx
Substitute to replace EVERY
x and dx:

 u 
  u 
  u 
4 x  1 dx 
12
dx
12 1
4
12
1
4
 
1
4
1
1 2 1
Solve
for dx
Find C:
du
du
1 2 1
u
C
 16 u 3 2  C

1
6
 4 x  1
32
C
F  x 
1
6
4
1
6
 4 x  1
32
C
 4  0.5  1
32
C
4  16  C
23
6  C
F  x 
1
6
 4 x  1
32
 236
Example 6
5 8
Evaluate
Define u
and du:

4
cos  2x  dx  
x 5 8
x  4
u  2x
cos  2 x  dx
Change the
Limits:
du  2  dx
Substitute:
5 8

4
cos  2x  dx 


5 4

2
Path 1:
cos u
5 4
1
2  2

1
2

x  4 x  58
2x  2 2x  54
u  2 u  54
du
2
1
2
cos u du
5 4
sin u 
2

 sin 54  sin 2   
Path 2:
1
2
2
4
 12
sin  2x  

5 8
 4
1
2

 sin  2    sin  2   
5
8
2
4
 12

4
Example 7
Evaluate
 4
0
Define u
and du:
tan x sec x dx  
2
x 0
u  tan x
du  sec x  dx
Substitute:

0
tan x sec 2 x dx 

tan x sec 2 x dx

x

x0
4
tan x  tan 0 tan x  tan 4
u0
u 1
Change the
Limits:
2
 4
x  4

1
0
Path 1:
u du
1
11
u
 u
1
2
1
2
11 1
2 1
0
0
12  12  02 
Path 2:
1
2
2
tan x
 12 tan

1
2
1
2
 4
0
2 
4
 12 tan 2 0
White Board Challenge
Evaluate:

5
0
x 25  x dx
2
125
3
Integration of Even and Odd Functions
Let f be integrable on the interval [-a,a].
1. If f is an even function, then

a
a
f  x  dx  2  f  x  dx
a
0
-a
a
Integration of Even and Odd Functions
Let f be integrable on the interval [-a,a].
2. If f is an odd function, then

a
a
-a
f  x  dx  0
a